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Be $S^1$ the unit circle in the plane, that is, $S^1= \lbrace (x,y) : x^2+y^2=1 \rbrace$ with the subspace topology. Show that $S^1 - \lbrace (1,0)\rbrace$ is homeomorphic to the open interval $(0,1)$.

I have an idea. Using this theorem:

Let $f:(X, \tau_1) \rightarrow (Y, \tau_2)$ be a homeomorphism. Let $a \in X$, so that $X - \lbrace a \rbrace$ is a subspace of $X$ and has induced topology $\tau_3$. Also, $Y-\lbrace f(a)\rbrace$ is subspace of $Y$ and has induced topology $\tau_4$. Then $(X - \lbrace a \rbrace, \tau_3)$ is homeomorphic to $(Y-\lbrace f(a)\rbrace, \tau_4)$. (Morris, "Topology without tears", remark 4.3.6)

My idea is to make $X=S^1$ and $Y=(0,1) \cup ?$ so, with the respective topologies using the theorem, $S^1-\lbrace (1,0)\rbrace$ is homeomorphic to $(0,1)$. But I'm stuck in the part of defining $Y$ and find the homeomorphism $f$ between $X$ and $Y$.

PD. Sorry about my English, but it is not my native language.

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    $\begingroup$ The issue with applying the result you mention is that any space $Y$ that will work will be "complicated" (i.e. it will just be $(0,1) \cup \{p\}$ for some abstract point $p$, and the topology on $Y$ will take some effort to define). You cannot, for example, take $p$ to be in $\mathbb{R}$ and the topology on $Y$ to be the usual subspace topology inherited from $\mathbb{R}$. This is probably why you are getting stuck. It is easier to approach this problem directly. Look for a function $(0,1)$ to $S^1 \setminus \{(1,0)\}$ that might be a homeomorphism. (Hint: trig functions.) $\endgroup$ Apr 21, 2014 at 4:04
  • $\begingroup$ Oh, of course... Another idea just came to me. $\endgroup$
    – SirWeigel
    Apr 21, 2014 at 4:35

1 Answer 1

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Be $f: (0, 2\pi) \rightarrow S^1-\lbrace (1,0) \rbrace$ such that if $\phi \in (0, 2\pi)$, $f(\phi)=(\cos(\phi),\sin(\phi))$. This is a homeomorphism, and $(0, 2\pi)$ is homeomorphic to $(0,1)$.

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